The value of sin\(\frac{\pi}6\) is \(\frac 12\)
∴ The question becomes sin–1\((\frac 12)\)

The range of principal value of sin–1 is \([\frac{-\pi}2,\frac{\pi}2]
\) and sin\((\frac{π}6)\) = \(\frac 12\)
Therefore, the value of sin–1(sin\(\frac{\pi}6\)) is \(\frac{\pi}6.\)
sin–1(sin x) = x
Provided x ∈ \([\frac{-\pi}2,\frac{\pi}2]
\)
∴ we can write sin–1(sin\(\frac{\pi}6\)) = \(\frac{\pi}6\)